The displacement of the particle varies with time according to the relation $\mathrm{x}=\frac{\mathrm{k}}{\mathrm{b}}[1-{\mathrm{e}}^{-\mathrm{bt}}]$. Then the velocity of the particle is

(1) $k\left(e^{-bt}\right)$

(2) $\frac{k}{b^2{e}^{-bt}}$

(3) $kb{e}^{-bt}$

(4) None of these 

If the velocity of a particle is (10 + 2t²) m/s, then the average acceleration of the particle between 2 sec and 5 sec is:

(1) 2 m/s²

(2) 4 m/s²

(3) 12 m/s²

(4) 14 m/s²

A bullet moving with a velocity of 200 cm/s penetrates a wooden block and comes to rest after traversing 4 cm inside it. What velocity is needed for travelling distance of 9 cm in same block?

1. 100 cm/s

2. 136.2 cm/s

3. 300 cm/s

4. 250 cm/s

A thief is running away on a straight road in a jeep moving with a speed of \(9\) m/s. A policeman chases him on a motorcycle moving at a speed of \(10\) m/s. If the instantaneous separation of the jeep from the motorcycle is \(100\) m, how long will it take for the policeman to catch the thief?
1. \(1\) s
2. \(19\) s
3. \(90\) s
4. \(100\) s

A car A is travelling on a straight level road with a uniform speed of 60 km/h. It is followed by another car B which is moving with a speed of 70 km/h. When the distance between them is 2.5 km, the car B is given a deceleration of 20 km/h². After how much time will B catch up with A

(1) 1 hr

(2) 1/2 hr

(3) 1/4 hr

(4) 1/8 hr

The speed of a body moving with uniform acceleration is u. This speed is doubled while covering a distance S. When it covers an additional distance S, its speed would become

(1) $\sqrt{3}u$

(2) $\sqrt{5}u$

(3) $\sqrt{11}u$

(4) $\sqrt{7}u$

Two trains one of length 100 m and another of length 125 m, are moving in mutually opposite directions along parallel lines, meet each other, each with speed 10 m/s. If their acceleration are 0.3 m/s² and 0.2 m/s² respectively, then the time they take to pass each other will be

1.  5 s
2.  10 s
3.  15 s
4.  20 s

A body starts from rest with uniform acceleration. If its velocity after n second is v, then its displacement in the last two seconds is

(1) $\frac{2v(n+1)}{n}$

(2) $\frac{v(n+1)}{n}$

(3) $\frac{v(n-1)}{n}$

(4) $\frac{2v(n-1)}{n}$

A point starts moving in a straight line with a certain acceleration. At a time t after beginning of motion the acceleration suddenly becomes retardation of the same value. The time in which the point returns to the initial point is

(1) $\sqrt{2t}$

(2) $(2+\sqrt{2})t$

(3) $\frac{t}{\sqrt{2}}$

(4) Cannot be predicted unless acceleration is given